1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric Field due to Point Charges

  1. Nov 3, 2014 #1
    1. The problem statement, all variables and given/known data

    The following is on a practice exam I have been completing. In advance, it is part b I am struggling with.

    Two point charges Q1 = +5*10^(-6)C and -2*10^(-6)C are 50cm apart.
    a) Where along a line that passes through the two charges is the electric potential zero (apart
    from at r=infinity)?

    b) Determine the magnitude of the electric field at that point. Explain why the electric field is non-zero.


    2. Relevant equations
    E=kQ/r^2

    V=kQ/r

    E=dV/dr


    3. The attempt at a solution
    I got part A immediately using V=kQ/r and superposition of electric potential, finding it to be 0.36m (which matched the solution). It is part b that I am struggling with. I have as follows:

    E=kQ/r^2
    Using superposition of electric fields, this gives
    E = k(5/(0.36^2) + -2/(0.5-0.36)^2), where k=1/(4*pi*E_0). This gave me an electric field of -570367N/C; however, the solution said it was 1.2*10^6 N/C. I believe the issue was in my implementation of the -2C in the above equation; however, am not entirely sure why this is incorrect or what to do to fix it.

    As for the explanation, my belief is that because E is changing in potential with respective to distance, as long as the potential is not set at a constant 0 - that is, the potential is changing as distance changes - the electric field could be non-zero. I think I'm okay with this, but please do not hesitate to correct me if I'm wrong!

    I'd really appreciate some help in figuring this out - or even just in where I went wrong. Please let me know if I've forgotten anything or my working isn't clear.

    Thank you so much
     
  2. jcsd
  3. Nov 3, 2014 #2
    You are right there was something wrong with the use of the -2 x 10-6C. The equation should be E=k[((5x10-6)/0.362)+((2x10-6)/0.142)], because we are finding the magnitude of the electric field, so the plus and minus signs doesn't matter.
     
    Last edited: Nov 3, 2014
  4. Nov 3, 2014 #3
    So does that mean that if we were finding the electric field, not magnitude, I would be correct? Or would I still use E=k[((5x10-6)/0.362)+((2x10-6)/0.142)] , then choose direction afterwards?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Electric Field due to Point Charges
Loading...